Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Tunan Jia on December 01, 2018, 05:06:45 PM

can anyone help with this question?

attached is my scanned answer, please check. Hopefully it can help you

Consider $ f(z) = \frac{\sqrt z}{z^2 + 2z +5}$
$z^2 + 2z +5 \implies $z= 1+2i or 1i2.
Only z=1+2i is up.
Res(f,1+2i) = $ \frac{\sqrt {1+2i}}{(1+2i)(12i)} $= $\frac{\sqrt {1+2i}}{4i}$
We compute $ \sqrt{1+2i}$ = a+ib. There are two solutions, but we must choose only the one whose argument is hafl of the grgument of 1+2i.
$a^2$  $b^2$ = 1 ab = 1
a = $ \sqrt{\frac{\sqrt5 1}{2}} $ = $ \frac{1}{b} $
I = re( $2 \pi i \frac{a+ib}{4i})$ = $ \frac{\pi}{2}$ $ \sqrt{\frac{\sqrt5 1}{2}}$